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Tamiku [17]
3 years ago
5

Consider the following scenario. A car for which friction is not negligible accelerates from rest down a hill, running out of ga

soline after a short distance (see below). The driver lets the car coast farther down the hill, then up and over a small crest. He then coasts down that hill into a gas station, where he brakes to a stop and fills the tank with gasoline. Identify the forms of energy the car has, and how they are changed and transferred in this series of events.
Physics
1 answer:
drek231 [11]3 years ago
6 0

Answer:

1) U-> K +W

2) K -> W

Explanation:

In this exercise, care must be taken as they indicate that the friction force (rubbing) is not negligible.

1 part at the top of the hill the car has gravitational potential energy, which is transformed in a part into kinetic energy and another part into heat by the work of the friction force that opposes the movement.

2 part when the other hill rises it loses kinetic energy that is transformed into gravitational potential energy and part in heat due to the work of the friction force on this hill.

3rd part in the last descent all the gravitational potential energy is transformed into kinetic energy and the application of the brakes is transformed into heat due to the negative lock of the friction force and filled with gasoline that has chemical energy that can be transformed in the engine.

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What are two situational examples of unbalanced forces?
polet [3.4K]

An example of a balanced force is two cards leaning against each other and not falling over, or two football players blocking each other but neither overpowering the other. An example of an unbalanced force is two cards leaning on each other then falling over, or two football players blocking each other, then one tackles the other.

4 0
3 years ago
A -turn rectangular coil with length and width is in a region with its axis initially aligned to a horizontally directed uniform
madam [21]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The maximum emf is \epsilon_{max}= 26.8 V

The emf induced at t = 1.00 s is \epsilon = 24.1V

The maximum rate of change of magnetic flux is   \frac{d \o}{dt}|_{max}  =26.8V

Explanation:

    From the question we are told that

        The number of turns is N = 44 turns

          The length of the coil is  l = 15.0 cm = \frac{15}{100} = 0.15m

          The width of the coil is  w = 8.50 cm =\frac{8.50}{100} =0.085 m

          The magnetic field is  B = 745 \ mT

          The angular speed is w = 64.0 rad/s

Generally the induced emf is mathematically represented as

        \epsilon = \epsilon_{max} sin (wt)

 Where \epsilon_{max} is the maximum induced emf and this is mathematically represented as

            \epsilon_{max} = N\ B\ A\ w

Where \o is the magnetic flux

            N is the number of turns

             A is the area of the coil which is mathematically evaluated as

             A = l *w

        Substituting values

           A = 0.15 * 0.085

               = 0.01275m^2

substituting values into the equation for  maximum induced emf

         \epsilon_{max} = 44* 745 *10^{-3} * 0.01275 * 64.0

                 \epsilon_{max}= 26.8 V

 given that the time t = 1.0sec

substituting values into the equation for induced emf  \epsilon = \epsilon_{max} sin (wt)

      \epsilon = 26.8 sin (64 * 1)

        \epsilon = 24.1V

   The maximum induced emf can also be represented mathematically as

              \epsilon_{max} = \frac{d \o}{dt}|_{max}

  Where  \o is the magnetic flux and \frac{d \o}{dt}|_{max} is the maximum rate at which magnetic flux changes the value of the maximum rate of change of magnetic flux is

         \frac{d \o}{dt}|_{max}  =26.8V

8 0
3 years ago
A defibrillator is used during a heart attack to restore the heart to its normal beating pattern. A defibrillator passes 18 A of
miskamm [114]

Answer:

q = 0.036 C

Explanation:

Given that,

Current passes through a defibrillator, I = 18 A

Time, t = 2 ms

We need to find the charge moved during this time. We know that,

Electric current = charge/time

q=It

Put all the values,

q=18\times 0.002\\\\q=0.036\ C

So, 0.036 C of charge moves during this time.

8 0
2 years ago
The “Big Bang” is an example of what type of scientific statement? A. A law B.a hypothesis C. A theory D. An equation
Ugo [173]

Answer:

theory

Explanation:

that's why it's called the Big Bang Theory

6 0
2 years ago
Read 2 more answers
When beryllium-7 ions (m = 11.65 × 10-27 kg) pass through a mass spectrometer, a uniform magnetic field of 0.205 T curves their
AysviL [449]

Answer:

ratio =0.3075 T

Explanation:

The magnetic field B creates a force on a moving charge such that

F = qvB

Now this causes a centripetal acceleration

F =  = mv^2/r

 so

qvB = mv^2/r ...........(i)

B = mv/(rq)  ...............(ii)

If  accelerating potential V is  same and  then  kinetic energy equals the potential energy difference

\frac{1}{2} mv^2 = Vq

v = \sqrt{(2Vq/m)}      put these value in equation (ii)

B = m\frac{\sqrt{(2Vq/m)} }{rq}  

simplifying we get  

B =m \frac{(\sqrt{ 2Vm/q})}{r}

for same location r will be same in both case

B_{7} = \frac{ \sqrt{(m_{7})(2V/q) }}{r}      ..............(iii)

B_{10} = \frac{ \sqrt{(m_{10})(2V/q) }}{r}    ..........(iv)

 dividing (iv) and (iii) equation we get

\frac{B_{10}}{B_{7}}   =   \sqrt{\frac{m_{10}}{{m_7}} }

{B_{10}}  =  B_{7}  \sqrt{\frac{m_{10}}{{m_7}} }

B_{10}       = 0.2574T\sqrt{\frac{  (1.663x10^-26}{(1.165x10^-26)}

so on solving we get  

             =0.3075 T

5 0
3 years ago
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